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I am not sure what exactly the question wants me to do:

The function $f$ has the properties that $f(1) = 6$ and $f(2x + 1) = 3f(x)$ for every integer $x$. What is the value of $f(63)$?

What exactly does it mean when $f(1) = 6$? and what is its relation with the other functions?

Arctic Char
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RudyGoburt
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  • It means you have some function, when applied to the input $1$ it gives the output of $6$. If $f(x) = 3 x^2$, for instance, then $f(1) = 3$. If instead $f(x) = x+ 9$, then $f(1) = 10$. Get it? This is the basics of arithmatic. – David G. Stork Sep 06 '20 at 04:09
  • @DavidG.Stork That makes sense. In the past, every time I would see f(x), I would just substitute it with y. – RudyGoburt Sep 06 '20 at 04:12

2 Answers2

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$$f(1)=6$$ and $$f(2x+1)=3f(x)$$ so
$$f(2(1)+1)=f(3)=3f(1)=3\cdot 6=18$$ $$f(2(3)+1)=f(7)=3f(3)=3\cdot 18=54 $$ $$f(2(7)+1)=f(15)=3f(7)=3\cdot 54=162 $$ $$f(2(15)+1)=f(31)=3f(15)=3\cdot 162=486$$ $$f(2(31)+1)=f(63)=3f(31)=3\cdot 486=1458 $$

C Squared
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A function is a way of assigning outputs to inputs. When someone says $f(1)=6$, they are saying that the value of $f$ is 6 when the input is 1.

Then, the equation $f(2x+1)=3f(x)$ is saying that two values of $f$ are related. Which two? The values of any pair of inputs $x$ and $2x+1$.

For instance, if $x=1$, then $2x+1=3$. Since we know $f(1)=6$, the equation above is telling us $$f(3) = f(2\cdot 1 + 1)=3f(1)=18$$

Now that you know the value of the function when the input is 3, you can repeat this idea for $x=3$ to figure out different values of the function.

Ryan
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  • I understood everything until this part: f(3)=f(2⋅1+1)=3f(1)=18. How did you get that off knowing the fact that f(1) = 6? – RudyGoburt Sep 06 '20 at 04:18
  • The first 3, inside the $f$, comes from the 2x+1 just above. The value at input 2x+1 is related to the value at input x=1. – Ryan Sep 06 '20 at 14:22